Transcript Document

Chemical Kinetics
Chapter 13
Sem 2/2014
1
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Chemical Kinetics
13.1 The Rate of a Reaction
Thermodynamics – does a reaction take place?
Kinetics – how fast does a reaction proceed?
Reaction rate is the change in the concentration of a
reactant or a product with time (M/s).
A
B
D[A]
rate = Dt
D[A] = change in concentration of A over
time period Dt
D[B]
rate =
Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.
2
A
B
D[A]
rate = Dt
D[B]
rate =
Dt
3
Reaction Rates and Stoichiometry
2A
B
Two moles of A disappear for each mole of B that is formed.
1 D[A]
rate = 2 Dt
aA + bB
D[B]
rate =
Dt
cC + dD
1 D[A]
1 D[B]
1 D[C]
1 D[D]
rate = ==
=
a Dt
b Dt
c Dt
d Dt
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Example 13.1
Write the rate expressions for the following reactions in terms of
the disappearance of the reactants and the appearance of the
products:
Example 13.1
Strategy To express the rate of the reaction in terms of the
change in concentration of a reactant or product with time, we
need to use the proper sign (minus or plus) and the reciprocal
of the stoichiometric coefficient.
Solution
(a) Because each of the stoichiometric coefficients equals 1,
(b) Here the coefficients are 4, 5, 4, and 6, so
Example 13.2
Consider the reaction
Suppose that, at a particular moment during the reaction,
molecular oxygen is reacting at the rate of 0.024 M/s.
(a) At what rate is N2O5 being formed?
(b) At what rate is NO2 reacting?
Example 13.2
Strategy To calculate the rate of formation of N2O5 and
disappearance of NO2, we need to express the rate of the
reaction in terms of the stoichiometric coefficients as in
Example 13.1:
We are given
where the minus sign shows that the concentration of O2 is
decreasing with time.
Example 13.2
Solution
(a) From the preceding rate expression we have
Therefore
Example 13.2
(b) Here we have
so
13.2 The Rate Law
The rate law expresses the relationship of the rate of a reaction
to the rate constant and the concentrations of the reactants
raised to some powers.
aA + bB
cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
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F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x=1
Quadruple [ClO2] with [F2] constant
rate = k [F2][ClO2]
Rate quadruples
y=1
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Rate Laws
•
Rate laws are always determined experimentally.
•
Reaction order is always defined in terms of reactant
(not product) concentrations.
•
The order of a reactant is not related to the
stoichiometric coefficient of the reactant in the balanced
chemical equation.
F2 (g) + 2ClO2 (g)
2FClO2 (g)
rate = k [F2][ClO2] 1
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Example 13.3
The reaction of nitric oxide with hydrogen at 1280°C is
From the following data collected at this temperature, determine
(a) the rate law
(b) the rate constant
(c) the rate of the reaction when [NO] = 12.0 × 10−3 M and
[H2] = 6.0 × 10 −3 M
Example 13.3
Strategy We are given a set of concentration and reaction rate
data and asked to determine the rate law and the rate constant.
We assume that the rate law takes the form
rate = k[NO]x [H2]y
How do we use the data to determine x and y?
Once the orders of the reactants are known, we can calculate k
from any set of rate and concentrations.
Finally, the rate law enables us to calculate the rate at any
concentrations of NO and H2.
Example 13.3
Solution (a) Experiments 1 and 2 show that when we double
the concentration of NO at constant concentration of H2, the
rate quadruples. Taking the ratio of the rates from these two
experiments
Therefore,
or x = 2, that is, the reaction is second order in NO.
Example 13.3
Experiments 2 and 3 indicate that doubling [H2] at constant
[NO] doubles the rate. Here we write the ratio as
Therefore,
or y = 1, that is, the reaction is first order in H2. Hence the rate
law is given by
which shows that it is a (2 + 1) or third-order reaction overall.
Example 13.3
(b) The rate constant k can be calculated using the data from
any one of the experiments. Rearranging the rate law, we
get
The data from experiment 2 give us
Example 13.3
(c) Using the known rate constant and concentrations of NO
and H2, we write
Comment Note that the reaction is first order in H2, whereas
the stoichiometric coefficient for H2 in the balanced equation
is 2. The order of a reactant is not related to the stoichiometric
coefficient of the reactant in the overall balanced equation.
13.3 Relation b/w Reactant concentration and Time
(a) First-Order Reactions
A
k=
product
D[A]
rate = Dt
rate
M/s
=
= 1/s or s-1
M
[A]
[A] = [A]0e−kt
rate = k [A]
D[A]
= k [A]
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
ln[A] = ln[A]0 - kt
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Example 13.4
The conversion of cyclopropane to propene in the gas phase is
a first-order reaction with a rate constant of 6.7 × 10−4 s−1 at
500°C.
(a) If the initial concentration of cyclopropane was 0.25 M, what
is the concentration after 8.8 min?
(b) How long (in minutes) will it take for the concentration of
cyclopropane to decrease from 0.25 M to 0.15 M?
(c) How long (in minutes) will it take to convert 74 percent of the
starting material?
Example 13.4
Strategy The relationship between the concentrations of a
reactant at different times in a first-order reaction is given by
Equation (13.3) or (13.4).
In (a) we are given [A]0 = 0.25 M and asked for [A]t after
8.8 min.
In (b) we are asked to calculate the time it takes for
cyclopropane to decrease in concentration from 0.25 M
to 0.15 M.
No concentration values are given for (c). However, if initially
we have 100 percent of the compound and 74 percent has
reacted, then what is left must be (100% − 74%), or 26%. Thus,
the ratio of the percentages will be equal to the ratio of the
actual concentrations; that is, [A]t/[A]0 = 26%/100%, or
0.26/1.00.
Example 13.4
Solution (a) In applying Equation (13.4), we note that because
k is given in units of s−1, we must first convert 8.8 min to
seconds:
We write
Hence,
Note that in the ln [A]0 term, [A]0 is expressed as a
dimensionless quantity (0.25) because we cannot take the
logarithm of units.
Example 13.4
(b) Using Equation (13.3),
(c) From Equation (13.3),
Graphical Determination of k
2N2O5
4NO2 (g) + O2 (g)
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Example 13.5
The rate of decomposition of azomethane (C2H6N2) is studied
by monitoring the partial pressure of the reactant as a function
of time:
The data obtained at 300°C are shown in the following table:
Are these values consistent with first-order kinetics? If so,
determine the rate constant.
Example 13.5
Strategy To test for first-order kinetics, we consider the
integrated first-order rate law that has a linear form, which is
Equation (13.4)
If the reaction is first order, then a plot of ln [A]t versus t (y
versus x) will produce a straight line with a slope equal to 2k.
Note that the partial pressure of azomethane at any time is
directly proportional to its concentration in moles per liter
(PV = nRT, so P  n/V). Therefore, we substitute partial
pressure for concentration [Equation (13.5)]:
where P0 and Pt are the partial pressures of azomethane at t =
0 and t = t, respectively.
Example 13.5
Solution First we construct the following table of t versus ln Pt.
Figure 13.11, which is based on the data given in the table,
shows that a plot of ln Pt versus t yields a straight line, so the
reaction is indeed first order. The slope of the line is given by
Example 13.5
According to Equation (13.4), the slope is equal to −k, so
k = 2.55 × 10−3 s−1 .
First-Order Reactions
The half-life, t½, is the time required for the concentration of a
reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2
ln
t½ =
[A]0
[A]0/2
k
ln 2
0.693
=
=
k
k
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First-order reaction
A
product
# of
half-lives [A] = [A]0/n
1
2
2
4
3
8
4
16
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Example 13.6
The decomposition of ethane (C2H6) to methyl radicals is a firstorder reaction with a rate constant of 5.36 × 10−4 s−1 at 700°C:
Calculate the half-life of the reaction in minutes.
Example 13.6
Strategy To calculate the half-life of a first-order reaction, we
use Equation (13.6). A conversion is needed to express the
half-life in minutes.
Solution For a first-order reaction, we only need the rate
constant to calculate the half-life of the reaction. From
Equation (13.6)
Second-Order Reactions
A
product
D[A]
rate = Dt
rate
M/s
=
= 1/M•s
k=
2
2
M
[A]
1
1
=
+ kt
[A]
[A]0
rate = k [A]2
D[A]
= k [A]2
Dt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
1
t½ =
k[A]0
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Example 13.7
Iodine atoms combine to form molecular iodine in the gas
phase
This reaction follows second-order kinetics and has the high
rate constant 7.0 × 109/M · s at 23°C.
(a) If the initial concentration of I was 0.086 M, calculate the
concentration after 2.0 min.
(b) Calculate the half-life of the reaction if the initial
concentration of I is 0.60 M and if it is 0.42 M.
Example 13.7
Strategy
(a) The relationship between the concentrations of a reactant at
different times is given by the integrated rate law. Because
this is a second-order reaction, we use Equation (13.7).
(b) We are asked to calculate the half-life. The half-life for a
second-order reaction is given by Equation (13.8).
Solution (a) To calculate the concentration of a species at a
later time of a second−order reaction, we need the initial
concentration and the rate constant. Applying Equation (13.7)
Example 13.7
where [A]t is the concentration at t = 2.0 min. Solving the
equation, we get
This is such a low concentration that it is virtually undetectable.
The very large rate constant for the reaction means that nearly
all the I atoms combine after only 2.0 min of reaction time.
(b) We need Equation (13.8) for this part.
For [I]0 = 0.60 M
Example 13.7
For [I]0 = 0.42 M
Check These results confirm that the half-life of a secondorder reaction, unlike that of a first-order reaction, is not a
constant but depends on the initial concentration of the
reactant(s).
Does it make sense that a larger initial concentration should
have a shorter half-life?
Zero-Order Reactions
A
product
D[A]
rate = Dt
D[A]
=k
Dt
rate
= M/s
k=
0
[A]
[A] = [A]0 - kt
rate = k [A]0 = k
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
[A]0
t½ =
2k
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Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
0
Rate Law
rate = k
1
rate = k [A]
2
[A]2
rate = k
Concentration-Time
Equation
[A] = [A]0 - kt
ln[A] = ln[A]0 - kt
1
1
=
+ kt
[A]
[A]0
Half-Life
t½ =
[A]0
2k
t½ = ln 2
k
1
t½ =
k[A]0
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A+B
Exothermic Reaction
+
AB+
C+D
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of
energy required to initiate a chemical reaction.
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13.6 Catalysis
A catalyst is a substance that increases the rate of a
chemical reaction without itself being consumed.
Ea
Uncatalyzed
k  A e
(  E a / RT )
k
Catalyzed
ratecatalyzed > rateuncatalyzed
Ea′ < Ea
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In heterogeneous catalysis, the reactants and the catalysts
are in different phases.
•
Haber synthesis of ammonia
•
Ostwald process for the production of nitric acid
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Catalytic converters
In homogeneous catalysis, the reactants and the catalysts
are dispersed in a single phase, usually liquid.
•
Acid catalysis
•
Base catalysis
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Enzyme Catalysis
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Binding of Glucose to Hexokinase
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